Question

A particle of mass $m$ and charge $+q$ approaches from a very large distance towards a uniformly charged ring of radius $R$ and charge, mass same as that of particle, with initial velocity $v_0$ along the axis of the as shown in the figure. What is the closet distance of approach between the ring and the particle? Assume the space to be gravity free and frictionless.

• A. $\sqrt{\frac{q^4}{{\pi }^2{\epsilon }_0^2{m}^2{v}_0^2}-R^2}$
• B. $\sqrt{\frac{3q^4}{2{\pi }^2{\epsilon }_2^0{m}^2{v}_4^0}-R^2}$
• C. $\sqrt{\frac{m^2{v}_4^0}{2{\pi }^2{q}^4{\epsilon }_2^0}-R^2}$
• D. $\sqrt{\frac{q^4}{4{\pi }^2{\epsilon }_2^0{m}^2{v}_4^0}-R^2}$

Answer 1

Answer: (A)

$\sqrt{\frac{q^4}{{\pi }^2{\epsilon }_0^2{m}^2{v}_0^2}-R^2}$

A mass of particles $=m$

Charge $=+q$

radius $=R$

Velocity $=v_0$

$d=?$

Now, Using Coulomb's law

$F=\frac{1}{4\pi {ϵ}_0}\frac{a^2}{r^2}-R^2$

or, $F^2=\frac{q^n}{{\pi }^2{ϵ}_0^2{v}^2}-R^2$

or, $\left(ma\right)=\sqrt{\frac{q^4}{{\pi }^2{ϵ}_0^2{v}_0^2{m}^2}-R^2}$